I want to find $A$ such that $$A\sum \limits_{n=m}^{\infty}1/n^{3}=1$$
for any natural value of $m$.
Convergence of $\sum \limits_{n=m}^{\infty}1/n^{3}$
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$\begingroup$
sequences-and-series
convergence
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7$A = \frac{1}{\sum_{k=m}^{\infty} \frac{1}{n^3}}$ there you go :) – 2017-02-17
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0So $A$ is to be solved in terms of $m$? – 2017-02-17
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0It is just $\zeta(3)=1.202$ minus finitely many terms. – 2017-02-17
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0Yes @SimplyBeautifulArt – 2017-02-17
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0I know that for big values of $m$ the solution can be $A=2m^{2}$, but for minor values there's no solution that does not involve a many terms sum? – 2017-02-17
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1No, the solution can never be $A=2m^2$, since $2m^2$ is rational, whereas $A$ is irrational. – 2017-02-17
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0@DietrichBurde, sorry, I meant that this solution is an approximation – 2017-02-17
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0$\large -\,{2 \over \Psi''\left(m\right)}$. $\Psi$: Digamma Function. – 2017-02-18
2 Answers
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In terms of the Hurwitz zeta function:
$$A=\frac1{\zeta(3,m)}$$
There is no further closed form, unless you allow
$$A=\frac1{\zeta(3)-\sum_{n=1}^{m-1}\frac1{n^3}}$$
Approximations may be done with the Euler-Maclaurin formula:
$$\sum_{n=1}^m\frac1{n^3}\approx\zeta(3)+\frac1{2m^2}+\frac1{2m^3}+\dots$$
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0Thanks, I didn't know the Hurwtiz zeta function. But you meant ζ(3,m), right? – 2017-02-17
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0Yes, that was my bad. – 2017-02-17
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If you're interested in approximations,
$$A = 2m^2 - 2m + 1 - \frac{1}{6 m^2} - \frac{1}{6 m^3} + \frac{1}{12 m^4} + \frac{1}{3m^5} + O\left(\frac{1}{m^6}\right) \ \text{as}\ m \to \infty $$